Question: What is the extraneous solution to these equations? $\dfrac{x^2 + 1}{x + 8} = \dfrac{-3x - 1}{x + 8}$
Answer: Multiply both sides by $x + 8$ $ \dfrac{x^2 + 1}{x + 8} (x + 8) = \dfrac{-3x - 1}{x + 8} (x + 8)$ $ x^2 + 1 = -3x - 1$ Subtract $-3x - 1$ from both sides: $ x^2 + 1 - (-3x - 1) = -3x - 1 - (-3x - 1)$ $ x^2 + 1 + 3x + 1 = 0$ $ x^2 + 2 + 3x = 0$ Factor the expression: $ (x + 2)(x + 1) = 0$ Therefore $x = -2$ or $x = -1$ The original expression is defined at $x = -2$ and $x = -1$, so there are no extraneous solutions.